let vector A=2i+3j-k which is anti parallel with vector B which has magnitude 6. find vector B

Let B = b₁ i + b₂ j + b₃ k.

Since A and B are anti-parallel, the angle between them is 180°. Then the dot product of A and B is

A • B = ||A|| ||B|| cos(180°)

⇒   A • B = - ||A|| ||B||

The magnitude of A is

||A|| = \(\sqrt{(2^2+3^2+(-1)^2)}\) = √14

so that

A • B = 2b₁ + 3b₂ - b₃ = -6√14

Because A and B are anti-parallel, their cross product is the zero vector.

A x B = (3b₃ + b₂) i - (2b₃ + b₁) j + (2b₂ - 3b₁) k = 0 i + 0 j + 0 k

⇒   3b₃ + b₂ = 0, 2b₃ + b₁ = 0 and 2b₂ - 3b₁ = 0

Now,

2b₂ - 3b₁ = 0 ⇒   b₂ = 3/2 b₁

2b₃ + b₁ = 0 ⇒   b₃ = -1/2 b₁

so that

2b₁ + 3/2 b₁ - (-1/2 b₁) = -6√14

7b₁ = -6√14

b₁ = -6/7 √14 = -6 √(2/7)

Similarly,

2b₂ - 3b₁ = 0 ⇒  b₁ = 2/3 b₂

3b₃ + b₂ = 0 ⇒   b₃ = -1/3 b₂

so that

2 (2/3 b₂) + 3b₂ - (-1/3 b₂) = -6√14

14/3 b₂ = -6√14

b₂ = -18/14 √14 = -9 √(2/7)

Finally,

2 (-6 √(2/7)) + 3 (-9 √(2/7)) - b₃ = -6√14

-39 √(2/7) - b₃ = -6√14

b₃ = 6√14 - 39 √(2/7) = 3 √(2/7)

and so the vector B is

B = -6 √(2/7) i - 9 √(2/7) j + 3 √(2/7) k

or equilvalently,

B = -3 √(2/7) (2 i + 3 j - k)

B = -3 √(2/7) A